﻿ The Equations for the Binding Energies of Nuclides in Terms of the Numbers of Protons and Neutrons They Contain
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The Equations for the Binding
Energies of Nuclides in Terms
of the Numbers of Protons
and Neutrons They Contain

## Introduction

Nuclei are composed of protons and neutrons. The term nucleon was coined to denote either a proton or a neutron. The protons and neutrons can form spin pairs. This is where two nucleons of opposite spin align like two bar magnets with opposite poles in contact. In fact the basis for a spin pair is the magnetic fields generated by the spinning electrical charges of the nucleons. (Although a neutron is electrically neutral it does have a radial distribution of charge which generates a net magnetic moment.)

There are three types of spin pairs: proton-proton, neutron-neutron and proton-neutron.

## Binding Energy

The mass of a nucleus is less than the sum of the masses of the protons and neutons it is composed of. The difference is called its mass deficit. The mass deficit expressed in energy units is called its binding energy. It is an amount of energy that must be supplied to break the nucleus up into its component parts.

The binding energy (BE) of a nuclide is composed of two parts. That which is due to the number and types of spin pairs it contains (BESP) and that due to the interactions of its nucleons (BENI); i.e,

#### BE(p, n) = BESP(p, n) + BENI(p, n)

Strictly speaking the following analysis could only be expected to apply for increments taking place within the same nuclear shell.

The number of spin pairs of protons which form is denoted as (p%2), p divided by 2 and rounded downward. The number of spin pairs of neutrons is likewise (n%2). The number of proton-neutron spin pairs is equal to the minimum of p and n.

Let Spp, Snn and Spn denote the binding energy created in the formation of the three types of spin pairs. Therefore

#### BESP(p, n) = Spp (p%2) + Snn(n%2) + Spnmin(p, n)

Spin pair formation is exclusive. One proton can form a spin pair with one other proton and one neutron and no more. The same applies for a neutron.

The binding energy created due to the interaction of nucleons is not exclusive. It is proportional to the interactions of the various types. The number of interactions of p protons with each other is p(p−1)/2. Likewise the number of interactions of n neutrons is n(n−1)/2. The number of interactions of p protons with n neutrons is pn.

Let Cpp be the binding energy created as a result of a proton-proton interaction. Likewise for Cnn and Cpn. Thus

#### BENI(p, n) = Cpp[p(p−1)/2] + Cnn[n(n−1)/2] + Cpn[pn]

The signs of the coefficients indicate the nature of force between the nucleons. If positive the force is an attraction; if negative it is a repulsion.

## Incremental Binding EnergiesFirst Differences

The incremental binding energies are the binding energy quantities less the value with one less nucleon. For example, the incremental binding energy of a neutron is

#### IBEn(p, n) = ΔnBE(p, n) = BE(p, n) − BE(p, n−1)

Here is the graph of the data for the case of the isotopes of Krypton (proton number 36).

The data on the incremental binding energies of protons for all nuclides with n=24 as a function of p shows a similar pattern.

These example show incremental binding energy decreasing with increasing numbers of nucleons of the same type. They are illustrations of like nucleons being repelled from each other.

If the incremental binding energy of nucleons are plotted against the number of nucleons of the opposite type it is an entirely different picture.

The rise in binding energy with increasing numbers of nucleons of the opposite type illustrates the case of opposite types of nucleons being attracted to each other.

The above graphs are just illustrations but exhaustive displays are available at neutrons, protons and neutron-proton pairs that like nucleons are repelled from each other and unlike attracted. In these displays the odd-even fluctuations are eliminated by making the units of analysis the spin pairs of nucleons rather than the individual nucleons. The theoretical analysis for the proposition is given in Interactions.

## Nucleonic Charge

After having made the case for like nucleons repelling each other and unlike ones being attracted the next step is explaining why this occurs. The obvious answer is that nucleons possess a nucleonic charge.

The character of the interaction of two nucleons can be represented by their possessing a nucleonic charge. If the nucleonic charges of two particles are Ω1 and Ω2 then their interaction is proportional to the product Ω1Ω2. Thus if the charges are of the same sign then they repel each other. If their charges are of opposite sign then they are attracted to each other.

The electrostatic repulsion between protons simply adds to the effective charge of protons. The amount of the addition depends upon the distance separating the protons. There is no qualitative change in the characteristics of a nucleus due to this force.

To evaluate the various incremental quantities it is convenient to define two functions. The even function ε(k) is defined as

#### ε(k) = 1 if k is even ε(k) = 0 otherwise

The negative step function σ(x) is defined as

Then

#### IBESPp(p, n) = ΔpBESP(p, n) = BESP(p, n) − BESP(p−1, n) = Sppε(p) + Spnσ(p−n) IBESPn(p, n) = ΔnBESP(p, n) = BESP(p, n) − BESP(p, n−1) = Snnε(n) + Spnσ(n−p)

The terms involving the even function ε( ) represent the sawtooth pattern.

For the nucleonic interaction incremental functions

#### IBNIp(p, n) = ΔpBENI(p, n) = BENI(p, n) − BENI(p−1, n) = Cpp[p − 1/2] + Cpnn IBNIn(p, n) = ΔnBENI(p, n) = BENI(p, n) − BENI(p, n−1) = Cnn[n − 1/2] + Cpnp

Note that ΔpBENI(p, n) with p held constant is a linear function of n. Likewise ΔnBENI(p, n) with n held constant is a linear function of p.

The slope of the line of ΔpBENI(p, n) versus n is Cpn. Thus if the slope of the line of ΔpBENI(p, n) versus n is positive then that is empirical evidence that a proton and a neutron are attracted to each other. On the other hand the slope of the line of ΔpBENI(p, n) versus p is Cpp. If that is negative it is evidence that through the interactive force protons are repelled from each other.

The slopes of the incremental binding energies of nucleons are basically the second differences of binding energies.

## Second Differences

One set of second differences in binding energies is

There are also

#### Δp,n2BE(p, n) = ΔpBE(p, n) − ΔpBE(p, n−1) Δn,p2BE(p, n) =ΔnBE(p, n) − ΔnBE(p−1, n)

These two quantities are generally equal. For the binding energy function given above they are both equal to Cpn

On the other hand

#### Δp,p2BE(p, n) = Cppand Δn,n2BE(p, n) = Cnn

The issue of the equality of the cross differences is investigated in Cross Differences.

## Testing the Model

The equation for binding energy

#### BE(p, n) = Spp (p%2) + Snn(n%2) + Spnmin(p, n) + Cpp[p(p−1)/2] + Cnn[n(n−1)/2] + Cpn[pn]

can be tested using regression analysis on the binding energies of 2931 nuclides. See Nucleus. There it is found that a regression model based upon a slightly modified version of the above equation explains 99.99 percent of the variation in the binding energy of almost three thousand nuclides. If the shell structure of nuclides is taken into account that figure is raised to 99.995 percent. The analysis concludes that nuclei are held together largely by the spin pairing of nucleons. That spin pairing is attractive and strong but exclusive in the sense that one nucleon can pair with one other nucleon of the same type and one of the opposite type and no more. Consequently nuclear shells consist of rings of modules of the form -N-P-P-N- (or equivalently -P-N-N-P-). These turn simultaneously in four different modes at such fantastically high rates that the dynamic appearance of a nucleous is that of concentric spherical shells. The cumulative occupancy numbers of the shells, called Nuclear Magic Numbers can be explained by a simple algorithm and abide by the formula

#### s(s² + 5)/3

where s is the maximum filled shell number. For example the the number of protons filling up the first 4 shells is 4(4²+5)/3=28. The number of neutrons filling up the first 5 shells is 5(5²+5)/3=50.

The binding energies of nuclides are also affected by a force due the interaction of their nucleons. This force is much weaker than that due to spin pairing but it is not exclusive. Its magnitudes are explained by a proton and a neutron having nucleonic charges. If the nucleonic charge of a proton is taken to be +1 then that of a neutron is −2/3. See Nucleus for the derivation.

Those figures for nucleonic charge explain why like nucleons are repelled from each other due to nucleonic interaction and unlike nucleons are attracted to each other. Those figures also explain why stable heavier nuclides have about fifty percent more neutrons than protons.

## Conclusion

The equation for the binding energy of a nuclide with p protons and n neutrons is

#### BE(p, n) = Spp (p%2) + Snn(n%2) + Spnmin(p, n) + Cpp[p(p−1)/2] + Cnn[n(n−1)/2] + Cpn[pn]

where (p%2) and (n%2) are the numbers of pairs among p protons and n neutrons, respectively. The coefficients Sxy and Cxy are the binding energies associated with the formation of a spin pair xy and an interaction between an x and a y, respectively.

The incremental binding energy functions are

#### ΔpBE(p, n) = Sppε(p) + Spnσ(p−n) + Cpp[p − 1/2] + Cpnn ΔnBE(p, n) = Snnε(n) + Spnσ(n−p) + Cnn[n − 1/2] + Cpnp

where the even function ε(k)=1 if k is even but zero otherwise.. The negative step function σ(k)=1 if k is negative but zero otherwise.

The second difference functions are:

#### Δp,p2BE(p, n) = Cpp Δn,n2BE(p, n) = Cnn Δp,n2BE(p, n) = Cpn Δn,p2BE(p, n) = Cpn

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These last two quantities are called the cross differences. They can be computed independently, but, as indicated, they should be equal. This proposition can be tested with the binding energy data and is found to hold. See Cross Differences.